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Projectile range formula

Range is how far a projectile lands from where it was thrown, measured horizontally. On level ground, with no air resistance, it is:

$$R = \frac{v_0^2 \sin 2\theta_0}{g}$$

$v_0$ is the launch speed, $\theta_0$ is the launch angle above the horizontal, and $g$ is 9.8 m/s$^2$. OpenStax University Physics prints this as equation 4.26 and states the condition it is derived under, which is that the projectile lands at the same height it was launched from.

That condition is doing a lot of work, and most of the wrong answers people get with this formula come from applying it when the condition fails.

Where it comes from

Split the launch velocity into components. Horizontal is $v_x = v_0\cos\theta_0$ and vertical is $v_y = v_0\sin\theta_0$. Nothing pushes horizontally, so $v_x$ never changes. Gravity acts vertically, so $v_y$ falls at $g$ per second.

The projectile is in the air until it comes back to launch height. Its vertical displacement is zero at that moment, so $0 = v_0\sin\theta_0\,t - \tfrac{1}{2}gt^2$. One solution is $t = 0$, which is the launch. The other is the landing:

$$t = \frac{2v_0\sin\theta_0}{g}$$

Multiply the time of flight by the constant horizontal speed:

$$R = v_0\cos\theta_0 \times \frac{2v_0\sin\theta_0}{g} = \frac{2v_0^2\sin\theta_0\cos\theta_0}{g}$$

The double-angle identity $2\sin\theta\cos\theta = \sin 2\theta$ turns that into the range formula. The whole derivation is two kinematic equations and one trigonometric identity.

The 45 degree result, and the pair of angles

$\sin 2\theta_0$ is largest when $2\theta_0 = 90^\circ$, so the range peaks at $\theta_0 = 45^\circ$. That is the only place the maximum can be, and it does not depend on the launch speed or on $g$. Fire the same shot on the moon and 45 degrees is still the answer, just with a much longer range.

The identity also says something less obvious. $\sin 2\theta$ is symmetric about $90^\circ$, so $\sin 60^\circ$ and $\sin 120^\circ$ are the same number. Two launch angles that add to $90^\circ$ give the same range. Thirty degrees and sixty degrees land in the same place. The steep shot spends longer in the air and moves slower horizontally, and the two effects cancel exactly.

Take a ball launched from the ground at $v_0 = 24$ m/s and $\theta_0 = 30^\circ$. Then $2\theta_0 = 60^\circ$ and $\sin 60^\circ = 0.866025$:

$$R = \frac{24^2 \times 0.866025}{9.8} = \frac{576 \times 0.866025}{9.8} = \frac{498.83}{9.8} = 50.9\ \text{m}$$

Check it the long way. $v_x = 24\cos 30^\circ = 20.78$ m/s and $v_y = 24 \sin 30^\circ = 12.0$ m/s. Time of flight is $2 \times 12.0 / 9.8 = 2.449$ s. Horizontal distance is $20.78 \times 2.449 = 50.9$ m. Same answer, which it has to be, because the formula is that calculation with the algebra done in advance.

Launch the same ball at $60^\circ$ and the range is 50.9 m again. Launch it at $45^\circ$ and you get $576 \times 1 / 9.8 = 58.8$ m, the most that a 24 m/s launch can ever do.

Launching from a height

Throw from a cliff, a table, or shoulder height, and the level-ground formula is wrong. It is wrong in a predictable direction, because the projectile is in the air longer than the formula assumes, so the true range is larger.

Go back to the vertical equation and stop assuming the displacement is zero. With the launch point a height $h$ above the landing point, taking down as negative, the landing satisfies $-h = v_0\sin\theta_0\,t - \tfrac{1}{2}gt^2$. That is a quadratic in $t$, and the positive root is:

$$t = \frac{v_0\sin\theta_0 + \sqrt{(v_0\sin\theta_0)^2 + 2gh}}{g}$$

Then the range is $v_0\cos\theta_0$ times that time, as before.

Same ball, 24 m/s at $30^\circ$, thrown from a cliff 15 m high. $v_y = 12.0$ m/s and $v_x = 20.78$ m/s.

$$t = \frac{12.0 + \sqrt{144 + 2 \times 9.8 \times 15}}{9.8} = \frac{12.0 + \sqrt{438}}{9.8} = \frac{12.0 + 20.928}{9.8} = 3.360\ \text{s}$$

$$R = 20.78 \times 3.360 = 69.8\ \text{m}$$

Nearly 19 m further than the level-ground answer, a 37 percent gain bought with 15 m of height. Set $h = 0$ in that expression and the square root becomes $v_0\sin\theta_0$, the numerator becomes $2v_0\sin\theta_0$, and the level-ground time of flight comes back. The general case contains the special one.

The best angle moves too. Once you launch from above the landing point, $45^\circ$ is no longer optimal and the peak drops below it. For this ball off this cliff the maximum range is at $39.1^\circ$, worth 72.2 m. The extra falling time is worth more when it is spent moving fast horizontally, and a flatter launch buys more horizontal speed.

Every one of these simulations integrates the trajectory numerically, so you can change the angle or the height and see whether the formula still holds.

References